This may be rather difficult question!!! who ever answerer's it must be extreemly smart!!!A small plastic ball with a mass of 6.50 x 10^-3 kg and with a charge of +0.150 uC is suspended from an insulating thread and hangs between the plates of a capacitor .The ball is in equilibrium, with the thread making an angle of 30.0 degrees with respect to the vertical. the area of each plate is 0.0150 m^2. what is the magnitude of the charge on each plate?

3. The attempt at a solution

I tried to find q using Q = mg/E, but it ended up in a mess since I was looking for the magnitude. The area is throwing me off...I don't know how to utilize it. Honestly, this is a very tough question! would love it if you answer it...... i now know the answer....but will give you it...only after you die of nerve!!!hahahaa!!

AlphaNumeric

3rd September 2007 - 08:16 AM

Suppose the charge density of the plates is σ and -σ respectively. Given you know their area, A, can you compute the electric field produced by each plate alone? Using superposition you can then combine the fields of the two plates to compute the field felt by the charge.

You know the angle the charge's teather makes with the vertical, so you can compute the vertical and horizontal components of the force on the charged mass.

You know that the horizontal force due to the two plates is equal to the horizontal component of the force on the charge due to the teether and gravity. From this, you can set up an equation which relates σ to the charge and mass of the object. Rearrange and solve.

Is this sufficent or do you need more specific pointers?

einstienear

3rd September 2007 - 09:32 AM

So you may have a point....but do you want to have more time before i give you the answer??

fleem

3rd September 2007 - 11:52 AM

You can't figure this out because you don't know how far apart the plates are, nor do you know the shape of the plates.

AlphaNumeric

3rd September 2007 - 01:33 PM

QUOTE (einstienear+Sep 3 2007, 10:32 AM)

So you may have a point....but do you want to have more time before i give you the answer??

I was under the impression you wanted help in how to solve it, hence why I gave a general overview in how to go about solving it. I know how to solve the system, I'm familiar with electromagnetics.

If you were trying to get people on the forum to solve it, you should have made that clearer, rather than giving the distinct impression you've been given this for homework and cant' work out how to do it.

QUOTE (Fleem+Sep 3 2007, 10:32 AM)

You can't figure this out because you don't know how far apart the plates are, nor do you know the shape of the plates.

Ignoring fringe effects from plate edges, the shape isn't important and the distance between plates isn't either because you can get by knowing the field is uniform between the plates. What's important is the charge on them.

fleem

3rd September 2007 - 03:11 PM

QUOTE (AlphaNumeric+Sep 3 2007, 07:33 AM)

Ignoring fringe effects from plate edges [...]

Well, it does depend on the precision desired, doesn't it.

I admit that if I had this on a test, I would be forced to assume a very low ratio of plate separation to plate area simply to come up with SOME answer. But I think I'd place a note after my answer or speak to the teacher afterward.

From the description of the problem, I don't believe there is much reason to think the ratio of plate separation to plate area is extremely low. I don't consider a practical interpretation of the word "capacitor" implying it--I've seen vacuum capacitors that were not much more than a couple of aluminum disks on sticks--and that's typically the sort of capacitor used in lab experiments like that described here. In fact, considering the density of most plastics, and thus the size of that plastic ball compared to the plate area, that ratio looks like it is quite high, here. Nor is a desired precision specified. Now I'm not saying we need to go so far as to consider the Casimir field or the extra space-time curvature contributed by the field's stress-energy tensor! But I am saying that using classical electrostatics, even a capacitor that has, for example, a ratio of plate separation to plate area of 1/20 will certainly need that considered to obtain a precision of a couple decimal places.

fleem

3rd September 2007 - 03:19 PM

Heh, I just thought of something anal (I would, wouldn't I)

That string holding the ball is at a 30 degree angle. For that string to fit between the plates, that places a lower limit on the ratio of plate separation to plate area that is QUITE high. (Yeah, OK, its lame, I know)

Confused2

3rd September 2007 - 10:31 PM

My 'take' on this would be to assume the field between the plates is uniform sowe get the force on the suspended charge from E = -V/d where d is the separation of the two plates from this we know V/d = k1

The capacitance C = permittivity * A /d= k2 / d

since V/d = k1

V = k1 d

Charge would then be Q = C V

hence Q = k1 d * k2/d

Edit .. I see AN said this a while back.

Farsight

5th September 2007 - 11:43 AM

Why don't you leave it with us eineinear? We'll write it all including the exact answer, all ready for the morning, then you can pick it up just before you head off to school.

fleem

5th September 2007 - 05:29 PM

QUOTE (Farsight+Sep 5 2007, 05:43 AM)

Why don't you leave it with us eineinear? We'll write it all including the exact answer, all ready for the morning, then you can pick it up just before you head off to school.

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